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Two smooth block are placed at a smooth ...

Two smooth block are placed at a smooth corner as shown in fig. Both the bloks are having mass m. We apply a force F on the block m. Block A presses block B in the normal direction, due to which pressing force on vertical wall will increase, and pressing force on the horizontal wall decreases, as we increases `F(theta = 37^(@)` with horizontal).

As soon as the pressing force on the horizontal wall by block B become zero, it will lose contact with ground. If the value of F further increases, block B will accelerate in the upward direction and simulaneously block A will towards right.
If the acceleration of block A is a rightwards, then the acceleration of block B will be

A

`(25)/(12) mg`

B

`(5)/(3) mg`

C

`(3)/(4) mg`

D

`(4)/(3) mg`

Text Solution

Verified by Experts

The correct Answer is:
C
For equilibrium of block A
`F = N sin theta`, `N = F // sin theta`
To lift block B from ground
`N cos theta ge m g` `implies (F)/(sin theta) cos theta ge mg`
`F ge m g tan theta = m g ((3)/(4))` So, `F_("min") = (3)/(4) mg`
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